Irrationality Measure of Pi 2

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14 Representations 20 14.1 Gamma Function Representation ...... 21 14.2 Zeta Function Representation ...... 21 14.3 Harmonic Kernel Representation ...... 21

15 Beta and Gamma Functions 21

16 Basic Diophantine Approximations Results 22 16.1 Rationals And Irrationals Numbers Criteria ...... 22 16.2 Irrationalities Measures ...... 23 16.3 NormalNumbers ...... 25

17 Problems 25 17.1 FlintHillClassSeries ...... 25 17.2 ComplexAnalysis ...... 26 17.3 IrrationalityMeasures ...... 27 17.4 PartialQuotients ...... 27 17.5 ConcatenatedSequences ...... 28 17.6 Exact Evaluations Of Power Series ...... 28 17.7 FlatHillClassSeries...... 28

18 Numerical Data 30 18.1 Data For The Irrationality Measure ...... 31 18.2 TheSineAsymptoticExpansions ...... 32 18.3 TheSineReﬂectionFormula...... 33

1 Introduction

Let α R be a real number. The irrationality measure µ(α) of the real number α is the ∈ inﬁmum of the subset of real numbers µ(α) 1 for which the Diophantine inequality ≥ p 1 α (1) − q ≪ qµ(α)

has ﬁnitely many rational solutions p and q. The analysis of the irrationality measure µ(π) 2 was initiated by Mahler in 1953, who proved that ≥ p 1 α > (2) − q q42

for all rational solutions p and q, see [18], et alii. This inequality has an effective constant for all rational approximations p/q. Over the last seven decades, the efforts of several authors have improved this estimate significantly, see Table 1. More recently, it was reduced to µ(π) 7.6063, see [26]. This note has the followings result. ≤ Theorem 1.1. For any number ε> 0, the Diophantine inequality

p 1 π (3) − q ≪ q2+ε

has ﬁnitely many rational solutions p and q. In particular, the irrationality measure µ(π)= 2. Irrationality Measure of Pi 3

Table 1: Historical Data For µ(π) Irrationality Measure Upper Bound Reference Year µ(π) 42 Mahler, [18] 1953 ≤ µ(π) 20.6 Mignotte, [19] 1974 ≤ µ(π) 14.65 Chudnovsky, [8] 1982 ≤ µ(π) 13.398 Hata, [13] 1993 ≤ µ(π) 7.6063 Salikhov, [26] 2008 ≤

After some preliminary preparations, the proof of Theorem 1.1 is assembled in Section 8. Three distinct proofs from three diﬀerent perspective are presented. Section 6 explores the relationship between the irrationality measure of π = [a0; a1, a2,...] and the magnitude of the partial quotients an. In Section 11 there is an application to the convergence of the Flint Hills series. A second independent proof of the convergence of this series also appears in Section 9. Some numerical data for the ratio sin pn/sin 1/pn is included in the last Section. These data match the theoretical result.

2 Notation

The set of natural numbers is denoted by N = 0, 1, 2, 3,... , the set of integers is de- { } noted by Z = . . . 3, 2, 1, 0, 1, 2, 3,... , the set of rational numbers is denoted by { − − − } Q = a/b : a, b Z , the set of real numbers is denoted by R = ( , ), and the set { ∈ } −∞ ∞ of complex numbers is denoted by C = x + iy : x,y R . For a pair of real valued or { ∈ } complex valued functions, f, g : C C, the proportional symbol f g is deﬁned by −→ ≍ c g f c g, where c , c R are constants. In addition, the symbol f g is deﬁned 0 ≤ ≤ 1 0 1 ∈ ≪ by f c g for some constant c> 0. | |≤ | |

3 Harmonic Summation Kernels

The harmonic summation kernels naturally arise in the partial sums of Fourier series and in the studies of convergences of continous functions. Definition 3.1. The Dirichlet kernel is defined by sin((2x + 1)z) (z)= ei2nz = , (4) Dx sin (z) −xX≤n≤x where x N is an integer and z R πZ is a real number. ∈ ∈ − Definition 3.2. The Fejer kernel is defined by

2 i2kz 1 sin((x + 1)z) x(z)= e = , (5) F 2 sin (z)2 0≤Xn≤x, −nX≤k≤n where x N is an integer and z R πZ is a real number. ∈ ∈ − These formulas are well known, see [16] and similar references. For z = kπ, the har- 6 monic summation kernels have the upper bounds (z) = (z) x , and (z) = |Kx | |Dx | ≪ | | |Kx | (z) x2 . |Fx | ≪ | | Irrationality Measure of Pi 4

An important property is the that a proper choice of the parameter x 1 can shifts ≥ the sporadic large value of the reciprocal sine function 1/sin z to (z), and the term Kx 1/sin(2x + 1)z remains bounded. This principle will be applied to the lacunary sequence p : n 1 , which maximize the reciprocal sine function 1/sin z, to obtain an eﬀective { n ≥ } upper bound of the function 1/sin z.

There are many different ways to prove an upper bound based on the harmonic summation kernels (z) and or (z). Three different approaches are provided below. Dx Fx 3.1 Real Sequence Technique The Dirichlet kernel in Definition 3.1 is a well defined continued function of two variables x, z R. Hence, for fixed z, it has an analytic continuation to all the real numbers x R. ∈ ∈ Accordingly, to estimate 1/ sin p , it is sufficient to fix z = p , and select a real number | n| n x R such that p x. ∈ n ≍ Theorem 3.1. Let z N be a large integer. Then, ∈ 1 z . (6) sin z ≪ | |

Proof. Let p /q : n 1 be the sequence of convergents of the real number π. Since { n n ≥ } the numerators sequence p : n 1 maximize the reciprocal sine function 1/sin z, it is { n ≥ } sufficient to prove it for z = pn. Define the associated sequence 22+2v2 + 1 x = πp , (7) n 22+2v2 n where v = v (p ) = max v : 2v p is the 2-adic valuation, and n 1. Replacing the 2 2 n { | n} ≥ parameters x = x R, z = p and applying Lemma 3.1 return n ∈ n sin ((2x + 1)z) = sin ((2x + 1)p ) (8) | | | n n | = cos (p ) |± n | = cos (p πq ) | n − n | 1, ≍ since the sequence of convergents satisfies p πq 0 as n . Lastly, putting them | n − n|→ →∞ all together, yield 1 (z) = Dx sin z sin((2x + 1)z)

1 n (p ) (9) ≪ |Dx n | sin((2 x + 1)p ) n n

xn 1 ≪ | | · z ≪ | | since z p r x, and the trivial estimate (z) x . | |≍ n ≍ n ≍ |Dx | ≪ | | Lemma 3.1. Let pn/qn : n 1 be the sequence of convergents of the real number π, and deﬁne the associated{ sequence≥ } 22+2v2 + 1 x = πp , (10) n 22+2v2 n where v = v (p ) = max v : 2v p is the 2-adic valuation, and n 1. Then 2 2 n { | n} ≥ Irrationality Measure of Pi 5

(i) sin (2x + 1)p )= cos p . n n ± n (ii) sin (2x + 1)p ) 1. | n n |≍ (iii) cos (2x + 1)p )= sin p . n n ± n 1 (iv) cos (2xn + 1)pn) . | |≪ qn Proof. (i) Routine calculations yield this:

sin((2xn + 1)pn) = sin(2xnpn + pn) (11)

= sin(2xnpn) cos(pn) + sin(pn) cos(2xnpn).

The ﬁrst sine function expression in the right side of (11) has the value

22+2v2 + 1 π sin(2x p ) = sin 2 πp2 = sin w = 1 (12) n n 22+2v2 n 2 n ± since 22+2v2 + 1 w = p2 (13) n 22v2 n is an odd integer. Similarly, the last cosine function expression in the right side of (11) has the value 22+2v2 + 1 π cos(2x p ) = cos 2 πz2 = cos w = 0. (14) n n 22+2v2 n 2 n Substituting (12) and (14) into (11) return

sin(2x + 1)p )= cos (p ) . (15) n n ± n (ii) It follows from the previous result:

sin(2x + 1)p ) = cos (p ) (16) | n n | |± n | = cos (p πq ) |± n − n | 1, ≍ since the sequence of convergents satisﬁes p πq 0 as n . (ii) The values in | n − n| → → ∞ (12) and (14), and routine calculations yield this:

cos((2xn + 1)pn) = cos(2xnpn + pn) (17)

= cos(2xnpn) cos(pn) + sin(pn) sin(2xnpn) = 1 sin (p ) . ± n (iv) This follows from the previous result:

cos((2x + 1)p ) = sin (p ) (18) | n n | |± n | = sin (p πq ) | n − n | 1 . ≪ qn Irrationality Measure of Pi 6

3.2 Integer Sequence Technique The analysis of the upper estimate of the reciprocal of the sine function 1/sin z using integer values x N is slightly longer than for real value x R presented in Theorem 3.1. ∈ ∈ However, it will be done here for comparative purpose and to demonstrate the technique. Theorem 3.2. Let z N be a large integer. Then, ∈ 1 z . (19) sin z ≪ | |

Proof. Let p /q : n 1 be the sequence of convergents of the real number π. Since { n n ≥ } the numerators sequence p : n 1 maximize the reciprocal sine function 1/sin z, it is { n ≥ } suﬃcient to prove it for z = pn. Deﬁne the associated sequence 22+2v2 + 1 x = πp , (20) n 22+2v2 n where v = v (p ) = max v : 2v p is the 2-adic valuation, and n 1. Replacing the 2 2 n { | n} ≥ integer part of the parameters x = [x ]+ x R and z = p yield n { n}∈ n sin ((2[x ] + 1)z) = sin ((2(x x ) + 1)p ) (21) | n | | n − { n} n | = sin ((2x + 1)p 2 x p ) | n n − { n} n | = sin ((2x + 1)p ) cos ( 2 x p ) + sin ( 2 x p ) cos ((2x + 1)p ) . | n n − { n} n − { n} n n n | Applying Lemma 3.1 return

sin ((2x + 1)z) = cos (p ) cos ( 2 x p ) + sin ( 2 x p ) ( sin (p ) (22) | n | |± n − { n} n − { n} n ± n | 1 1 1 cos ( 2 x p ) + sin ( 2 x p ) . ≫ − 2q2 · − { n} n − { n} n · q n n

Applying Lemma 3.2 return the lower bound

1 1 1 sin ((2x + 1)z) 1 1 1 (23) | n | ≫ − 2q2 · − 2q2 − · q n n n

1 . ≫ Lastly, putting them all together, yield 1 (z) = Dx sin z sin((2x + 1)z)

1 (p ) (24) ≪ D[xn] n sin((2[ x ] + 1)p ) n n

xn 1 ≪ | | · z ≪ | | since z p r x, and the trivial estimate x ](z) x . | |≍ n ≍ n ≍ D[ n ≪ | n|

Lemma 3.2. Let pn/qn : n 1 be the sequence of convergents of the real number π, and deﬁne the associated{ sequence≥ } 22+2v2 + 1 x = πp , (25) n 22+2v2 n where v = v (p ) = max v : 2v p is the 2-adic valuation, and n 1. Let x = 2 2 n { | n} ≥ || || min x n : n 1 denote the minimal distance to an integer. Then {| − | ≥ } Irrationality Measure of Pi 7

1 . (i) 1 2 cos (2 xn pn) − 2qn ≪ | || || | 1 . (ii) 1 2 cos (2 xn pn) − 2qn ≪ | { } | 2π2 2π4 . (iii) cos (2 xn pn) 1 12 + 24 | { } |≪ − qn 3qn Proof. (i) Let x denotes the fractional part, and let x = min x , 1 x . Clearly, { n} || || {{ n} −{ n}} it has the upper bound π x = r s (26) || n|| 2a n − n

≤ sn 1 , ≤ 2

2+2v2 (pn) where a =2+2v2(pn), rn = (2 + 1)pn, and sn = [xn] > 1 or sn = [xn] + 1 > 1 are integers as n . Hence, →∞ cos (2 x p ) cos (2(1/2)p ) (27) | || n|| n | ≫ | n | = cos (p ) | n | = cos (p πq ) | n − n | 1 1 2 . ≥ − 2qn (ii) If x = x , the lower bound || || { n} 1 cos (2 xn pn) = cos (2 xn pn) 1 2 (28) | { } | | || || |≫ − 2qn is trivial. If x = 1 x , the lower bound has the form || || − { n} 1 1 2 cos (2 xn pn) (29) − 2qn ≪ | || || | = cos(2(1 x )p ) | − { n} n | = cos( 2 x p ) cos(2p ) + sin( 2 x p ) sin(2p ) | − { n} n n − { n} n n | 2 cos(2 x p ) 1 sin(2 x p ) ≤ { n} n · − { n} n · q n

cos(2 xn pn) . ≪ | { } | (iii) Utilize the inequality 1/q7 p πq and p = πq + O(1/q ) as n . n ≤ | n − n| n n n →∞ Note that in general, for an irrational number α R Z, the least distance to an integer ∈ − has the form 1 xn = αrn sn , (30) || || | − |≤ sn where r /s α is the sequence of convergents. n n → Irrationality Measure of Pi 8

3.3 Continued Fraction Technique Theorem 3.3. Let ε> 0 be an arbitrary small number, and let z C πZ be a large real number. Then, ∈ − 1 z 1+ε . (31) sin z ≪ | |

Proof. Let z = kπ be a large number, where k Z, and let u /v : m 1 be the | | 6 ∈ { m m ≥ } sequence of convergents of the algebraic irrational number √α R, such that √α 1/π. ∈ ≤ Moreover, let v z be a large integer, and let x = απv for which m ≍ | | m v 1 αv2 u2 + m < , (32) m − m 2π 2π see Lemma 3.3. Then, substituting these parameters yield

sin(2x + 1)z) sin ((2απvm + 1) vm) (33) | | ≍ | 2 | = sin 2απvm + vm 2 2 = sin 2απvm 2πum + vm − v = sin 2π αv2 u2 + m m − m 2π v αv2 u2 + m ≫ m − m 2π 2 2 αvm tm , ≫ − where t2 = u2 v /2π u2 . Next, use the principle of best rational approximations, m m − m ∼ m see Lemma 16.5, to derive this estimate:

αv2 t2 = √αv t √αv + t (34) m − m m − m m m √αvm tm vm ≫ − √αvm um vm. ≫ −

To continue, observe that the irrationality measure µ (√α) = 2, and its partial quotients ε √α = [b0, b1,...] are bounded by bm = O(rm), where ε > 0 is a small arbitrary number, see Lemma 16.4, and Example 16.1. Hence, the previous expression satisﬁes the inequality 1 √ ε αvm um vm 1, (35) vm ≪ − ≪ where ε> 0 is an arbitrary small number. Merging (34) and (35) return 1 1 vε . (36) sin((2x + 1)z) ≪ √αv u v ≪ m | | | m − m| m Lastly, putting them all together, yield 1 (z) = Dx sin z sin((2x + 1)z)

1 (z) ≪ |D x | · sin((2 x + 1)z)

ε 2π√αvm vm (37) ≪ · v1+ε ≪ m z 1+ε . ≪ | | The last line uses the trivial estimate (z) x v z . |Dx | ≪ | |≍ m ≍ | | Irrationality Measure of Pi 9

Lemma 3.3. Let u /v : m 1 be the sequence of convergents of the algebraic irra- { m m ≥ } tional number √α R, such that √α 1/π. Then, as n , ∈ ≤ →∞ v 1 αv2 u2 + m < . (38) m − m 2π 2π

Proof. It is suﬃcient to show that the error incurred in the approximation of the irrational number α by the sequence (u2 t )/v2 : m 1 , where t2 = u2 (1 v /2πu2 ) u , { m − m m ≥ } m m − m m ≈ m is bounded by a constant: v αv2 u2 + m = √αv t √αv + t (39) m − m 2π m − m m m

2 √α √αvm tm vm ≤ − < 2√α 1 < , 2π where t u , and 2√α 1/2π. m ∼ m ≤ 4 A suitable choice to meet the condition in (38) is √α = 1/2√d, with 16π4 < d.

4 Elementary Techniques

A similar case arises for the sine function sin(απn as n . This is handles by the → ∞ observing that 1/qa−1 < p αq 1/q is small as n . Thus, the denominators n | n − n| ≤ n → ∞ sequence q : n 1 maximize the reciprocal sine function 1/sin z. Hence, it is suﬃcient { n ≥ } to consider the inﬁnite series over the denominators sequence q : n 1 . { n ≥ } Theorem 4.1. Let α R be an irrational number of irrationality measure µ(α) = a. Then ∈ 1 qa−1. (40) sin απq ≪ n | n| Proof. Let α = [a , a , a ,...] be the continued fraction of the number α, and let p /q : 0 1 2 { n n n 1 be the sequence of convergents. Then ≥ } 1 1 = (41) sin απq sin(πp απq ) | n| | n − n | 1 ≥ π pn αqn a−| 1 − | > qn . as n . →∞ 5 Asymptotic Expansions Of The Sine Function

Theorem 5.1. Let pn/qn be the sequence of convergents of the irrational number π. Then,

1 sin sin p (42) p ≍ n n as p . n →∞ Irrationality Measure of Pi 10

Lemma 5.1. Let pn/qn be the sequence of convergents of the irrational number π = [a0; a1, a2,...]. Then, the followings hold. (i) sin p = sin (p πq ), for all p and q . n n − n n n 1 1 1 1 1 1 (ii) sin = 1 + , as p . p p − 3! p2 5! p4 −··· n →∞ n n n n (iii) cos 2p = cos (2(p πq )), for all p and q . n n − n n n 6 Bounded Partial Quotients

A result for the partial quotients based on the properties of continued fractions and the asymptotic expansions of the sine function is derived below. Theorem 6.1. Let π = [a ; a , a ,...] be the continued fraction of the real number π R. 0 1 2 ∈ Then, the nth partial quotients a N are bounded. Speciﬁcally, a = O(1) for all n 1. n ∈ n ≥ Proof. Consider the real numbers 1 1 C = pn+1 πqn+1 and D = . (46) − − qn qn For all suﬃciently large integers n 1, the Dirichlet approximation theorem and the ≥ addition formula sin(C + D) = cos C sin D + cos D sin C lead to 1 pn+1 πqn+1 (47) qn+1 ≫ | − | sin (p πq ) ≫ | n+1 − n+1 | 1 1 = sin p πq + n+1 − n+1 − q q n n 1 1 = cos C sin + cos sin C . q q n n

Irrationality Measure of Pi 11

The sequence p : n 1 maximizes the cosine function and minimizes the sine function. { n ≥ } Hence, by Lemma 6.2, 2 C2 1 2 1 cos C 1, (48) − qn ≤ − 2 ≤ ≤ and 1 1 C3 2 3 C sin C . (49) 2qn − 48qn ≤ − 6 ≤ ≤ qn Substituting these estimates into the reverse triangle inequality X + Y X Y , pro- | |≥ || |−| || duces 1 1 1 cos C sin + cos sin C (50) q ≫ q q n+1 n n 1 1 1 1 (1) (1) ≥ q − 6q3 − 2q − 48q3 n n n n c 0 , ≥ q n where c0 > 0 is a constant. These show that 1 1 (51) qn+1 ≫ qn as n . Furthermore, (51) implies that q q = a q + q as claimed. →∞ n ≫ n+1 n+1 n n−1 Lemma 6.1. Let pn/qn be the sequence of convergents of the real number π = [a0; a1, a2,...]. Then, as n , →∞ 1 1 2 1 1 13 (i) p πq . (ii) p πq . 2q ≤ n+1 − n+1 − q ≤ q 2p ≤ n+1 − n+1 − q ≤ p n n n n n n

Proof. (i) Using the triangle inequality and Dirichlet approximat ion theorem yield the upper bound

1 1 p πq p πq + (52) n+1 − n+1 − q ≤ | n+1 − n+1| q n n c 1 0 + ≤ qn+1 qn 2 , ≤ qn since qn+1 = an+1qn + qn−1. The lower bound 1 1 p πq p πq (53) q − n+1 − n+1 ≥ q − | n+1 − n+1| n n

1 c1 ≥ qn − qn+1 1 , ≥ 2qn where c0 > 0 and c1 > 0 are small constants. (ii) Use the Hurwitz approximation theorem 1 1 πq p πq + (54) n 2 n n 2 − √5qn ≤ ≤ √5qn Irrationality Measure of Pi 12

to convert it to the upper bound in term of pn: 1 2 13 p πq . n+1 − n+1 − q ≤ q ≤ p n n n as n as claimed. →∞ Lemma 6.2. Let p /q : n 1 be the sequence of convergents of π, and let C = { n n ≥ } p πq q . Then, n+1 − n+1 − n 2 2 1 . (i) 1 2 cos C 1 2 + 4 − qn ≤ ≤ − qn 6qn 1 1 2 . (ii) 3 sin C 2qn − 48qn ≤ ≤ qn Proof. Use Lemma 6.1.

7 Statistics And Example

The result of Theorem 6.1 indicates that the real number π ﬁts the proﬁle of a random irrational number. The geometric mean value of the partial quotients of a random irrational number has the value

1/n

K0 = lim ak = 2.6854520010 ..., (55) n→∞   kY≤n   which is known as the Khinchin constant. In addition, the Gauss-Kuzmin distribution speciﬁes the frequency of each value by

1 p(a = k)= log 1 . (56) k − 2 − (k + 1)2

For a random irrational number, the proportion of partial quotients ak = 1, 2 is about p(a )+ p(a ) .5897. A large value a 106 is very rare. In fact, 1 2 ≤ k ≥ 1 p(a = 106)= log 1 = 0.00000000000144, (57) k − 2 − (106 + 1)2 and the sum of probabilities is

1 p(ak = k)= log2 1 2 0.000001. (58) 6 6 − − (k + 1) ≤ kX≥10 kX≥10

Example 7.1. A short survey of the partial quotients π = [a0; a1, a2,...] is provided here to sample this phenomenon, most computer algebra system can generate a few thousand terms within minutes. This limited numerical experiment established that a 21000 for n ≤ n 10000. The value a = 20776 is the only unusual partial quotient. This seems to be ≤ 432 an instance of the Strong Law of Small Number.

π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15, 3, 13, 1, 4, 2, (59) 6, 6, 99, 1, 2, 2, 6, 3, 5, 1, 1, 6, 8, 1, 7, 1, 2, 3, 7, 1, 2, 1, 1, 12, 1, 1, 1, 3, 1, 1, 8, 1, 1,...]. Irrationality Measure of Pi 13

8 Main Result

The analysis of the irrationality measure µ(π) of the number π provided here is completely independent of the prime number theorem, and it is not related to the earlier analysis used by many authors, based on the Laplace integral of a factorial like function

1 n! k+1 e−tzdz, (60) i2π z(z 1)(z 2)(z 3) (z n) ZC − − − · · · − where C is a curve around the simple poles of the integrand, and k 0, see [18], [19], [8], ≥ [13], [26], and [6] for an introduction to the rational approximations of π and the various proofs. The improvements made using rational integrals and the prime number theorem, as the new estimate in [29], are limited to incremental improvements, and no where near the numerical data, see Table 3.

8.1 First Proof The ﬁrst proof of the irrationality measure µ(π) of the number π is based on a result for the upper bound of the reciprocal sine function over the sequence of p : n 1 derived { n ≥ } in Section 3. This is equivalent to a result in Section 4.

Proof. (Theorem 1.1) Let ε> 0 be an arbitrary small number, and let p /q : n 1 be { n n ≥ } the sequence of convergents of the irrational number π. By Theorem 3.1, the reciprocal sine function has the upper bound

1 p1+ε. (61) sin p ≪ n n

Moreover, sin(p ) = sin (p αq) if and only if αq = πq , where q is an integer. These n n − n n information lead to the following relation. 1 1+ε sin (pn) (62) pn ≪ | | sin (p πq ) ≪ | n − n | p πq ≪ | n − n| for all suﬃciently large pn. The rational approximation pn πqn < 1/qn implies that −1 | − | πqn = pn + O qn . Therefore,

p 1 π n (63) − q ≫ 1+ε n qnpn

1 ≫ q2+ε 1 = . qµ(π)+ε

Clearly, this implies that the irrationality measure of the real number π is µ(π) = 2, see Deﬁnition 16.1. Quod erat demontrandum.

This theory is consistent with the numerical data in Table 3, which shows the measure approaching 2 as the rational approximation p /q π. n n → Irrationality Measure of Pi 14

8.2 Second Proof The second proof of the irrationality measure µ(π) of the number π is based on the asymptotic expansion of the sine function in Section 5. Related results based on the cosine and sine functions appears in [1] and [10].

Proof. (Theorem 1.1) Given a small number ε> 0, it will be shown that ψ(q) qµ(π)+ε ≤ ≤ q2+ε is the irrationality measure of the real number π.

Observe that in Theorem 5.1, the asymptotic expansion of the sine function satisﬁes 1 sin (p ) sin . (64) | n |≫ p n

2+ε This inequality leads to the irrationality measures ψ(q)= q for real number π. Specif- ically,

p πq sin (p πq ) (65) | n − n| ≫ | n − n | = sin (p ) | n | 1 sin ≫ p n

1 ≫ pn 1 ≫ qn 1 ≫ q1+ε

−1 since pn = πqn + O qn . Therefore,

ψ(q) qµ(π)+ε q2+ε. (66) ≤ ≤ Quod erat demontrandum.

8.3 Third Proof The third proof of the irrationality measure µ(π) of the number π is based on the Dio- phantine inequality 1 p 1 α n (67) 2q q ≤ − q ≤ µ(α) n+1 n n qn for irrational numbers α R of irrationality measure µ(α) 2, confer Lemma 16.4, Deﬁ- ∈ ≥ nition 16.1, and a result for the partial quotients in Theorem 6.1.

Proof. (Theorem 1.1) Take the logarithm of the Diophantine inequality associated to the real number π. Speciﬁcally, 1 p 1 π n (68) 2q q ≤ − q ≤ µ(π) n+1 n n qn

to reach log 2q q log 2q µ(π) n+1 n =1+ n+1 . (69) ≤ log qn log qn Irrationality Measure of Pi 15

By Theorem 6.1, there is a constant c> 0 for which a c. Hence, n ≤

qn+1 = an+1qn + qn+1 (70) 2a q ≤ n+1 n 2cq . ≤ n Substitute the last estimate (70) into (69) obtain

log 4cq µ(π) 1+ n (71) ≤ log qn log 4c 2+ ≤ log qn for all suﬃciently large n 1. Taking the limit yields ≥ log 4c µ(π) = lim 2+ = 2. (72) n→∞ log q n Quod erat demontrandum.

9 Convergence Of The Flint Hills Series I

The ﬁrst analysis of the the convergence of the Flint Hills series is based on the upper bound of the reciprocal sine function 1/sin n as n , in Theorem ??. This is equivalent →∞ to Theorem 3.1 for the upper bound of the harmonic summation kernel sin((2x + 1)z) (z)= ei2nz = . (73) Dx sin (z) −xX≤n≤x Let π = [a , a , a ,...] be the continued fraction of the number π, and let p /q : n 1 0 1 2 { n n ≥ } be the sequence of convergents. The diﬃculty in proving the convergence of the series (76) arises from the sporadic maximal values of the function 1 1 = (74) sin p sin(p πq ) | n| | n − n | 1 ≥ pn πqn | 42 − | > qn . at the integers z = p as n . The irrationality exponent 42 in the Diophantine n → ∞ inequality 1 p 1 π n (75) q42 ≤ − q ≤ q n n n was determined in [18]. Earlier questions on the convergence the series (76) appears in

[23, p. 59], [27, p. 583], [1], et alii.

Theorem 9.1. If the real numbers u > 0 and v > 0 satisfy the relation u v > 0, then the Flint Hills series − 1 (76) nu sinv n nX≥1 is absolutely convergent. Irrationality Measure of Pi 16

Proof. As 1/q41 < p πq 1/q is small as n , the numerators sequence p : n | n − n| ≤ n → ∞ { n n 1 maximize the reciprocal sine function 1/sin z. Hence, it is sufficient to consider ≥ } the infinite series over the numerators sequence p : n 1 . To accomplish this, write { n ≥ } the series as a sum of a convergent infinite series and the lacunary infinite series (over the sequence of numerators): 1 1 1 = + (77) nu sinv n mu sinv m pu sin (p )v n≥1 m≥1 n≥1 n n X mX6=pn X 1 c0 u v . ≤ pn sin (pn) nX≥1 where c0, c1, c2, c3, c4 > 0 are constants. By Theorem ??, the reciprocal of the sine function is bounded 1 c p . (78) sin(p ) ≤ 1 n n

Applying this bound yields

v 1 (c1pn) u v u pn(sin pn) ≤ pn nX≥1 nX≥1 1 c2 u−v . (79) ≤ pn nX≥1 By the Binet formula for quadratic recurrent sequences, the sequence pn = anpn−1 + pn−2, a 1, has exponential growth, namely, n ≥ n 1 1+ √5 pn (80) ≥ √5 2 ! for n 2. Now, replacing (80) into (79) returns ≥ 1 2 (u−v)n c2 u−v c3 pn ≤ 1+ √5 nX≥1 nX≥1 1 (u−v)n c . (81) ≤ 4 2 nX≥1 Hence, it immediately follows that the infinite series converges whenever u v > 0. − Example 9.1. The infinite series 1 (82) n3 sin2 n nX≥1 has u v = 3 2 = 1 > 0. Hence, by Theorem 9.1, it is convergent. − − Example 9.2. Let ε> 0 be an arbitrary small number. The infinite series 1 (83) n1+ε sin n nX≥1 has u v =1+ ε 1= ε> 0. Hence, by Theorem 9.1, it is absolutely convergent. − − Irrationality Measure of Pi 17

Table 2: Comparison Of The Partial Sums Px and Qx For (u, v) = (3, 2). x Px Qx 1 1.422829 1.422829 3 3.423233 1.887049 22 4.754112 3.085767 355 29.405625 27.683949

1 The Partial Sum Px = n≤x n3 sin2 n P 35 30 25 x

P 20 15 10 5 0 01232122 354355 400 500 x

10 The Flint Hills Series And The Lacunary Sine Series

A comparison of the partial sum of the Flint Hills series 1 P = (84) x nu sinv n nX≤x and the partial sum of the Lacunary Sine series 1 Qx = u v (85) pn sin (pn) pXn≤x respectively, is tabulated in Table 2. It demonstrates that the Lacunary Sine series infuses an overwhelming contribution to the complete sum.

11 Convergence Of The Flint Hills Series II

The second analysis of the the convergence of the Flint Hills series is based on the asymptotic relation 1 sin (p ) sin , (86) n ≍ p n where p /q Q is the sequence of convergents of the real number π R. In some way, n n ∈ ∈ the representation (86) removes any reference to the diﬃcult problem of estimating the maximal value of the function 1/sin n as n . Earlier study of the convergence the → ∞ series (87) appears in [27, p. 583], [1], et alii. Irrationality Measure of Pi 18

Theorem 11.1. If the real numbers u> 0 and v > 0 satisfy the relation u v > 0, then the Flint Hills series − 1 (87) nu sinv n nX≥1 is absolutely convergent.

Proof. Let p /q : n 1 be the sequence of convergents of the real number π. Since { n n ≥ } the numerators sequence p : n 1 maximize the reciprocal sine function 1/sin z. { n ≥ } Hence, it is sufficient to consider the infinite series over the lacunary numerators sequence p : n 1 . Substituting the numerators sequence returns { n ≥ } 1 1 u v u v (88) n sin n ≪ pn(sin pn) Xn≥1 Xn≥1 1 v , ≪ pu sin 1 Xn≥1 n pn see Theorem 5.1. Substituting the Taylor series at infinity, see Lemma 5.1, return 1 1 v = v 1 u u 1 1 1 1 1 n≥1 pn sin pn n≥1 p 1 + X X n p − 3! p2 5! p4 −··· n n n 1 u−v . (89) ≪ pn nX≥1

By the Binet formula for recurrent quadratic sequences, the sequence pn = anpn−1 + pn−2, a 1, has exponential growth, namely, n ≥ n 1 1+ √5 pn (90) ≥ √5 2 ! for n 2. Replacing (90) into (89) returns ≥ 1 2 (u−v)n u−v pn ≪ 1+ √5 nX≥1 nX≥1 1 (u−v)n . (91) ≪ 2 nX≥1 Hence, it immediately follows that the inﬁnite series converges whenever u v > 0. − 12 Convergence Of Some Flint Hills Type Series

The analysis for the convergence of some modified Flint Hills series as 1 (92) nu sinv απn Xn≥1 have many similarities to the of the Flint Hills series. Given an irrational number α R ∈ of irrationality measure µ(α) = a, let π = [a0, a1, a2,...] be the continued fraction of the number α, and let p /q : n 1 be the sequence of convergents. The difficulty in { n n ≥ } Irrationality Measure of Pi 19 proving the convergence of the series (76) arises from the sporadic maximal values of the function 1 1 = (93) sin απq sin(πp απq ) | n| | n − n | 1 ≥ π pn αqn a−| 1 − | > qn . as n . →∞ Theorem 12.1. If the real numbers u> 0 and v > 0 satisfy the relation u (a 1)v > 0, then the series − − 1 (94) nu sinv απn Xn≥1 is absolutely convergent. Proof. As 1/qa−1 < p αq 1/q is small as n , the numerators sequence n | n − n| ≤ n → ∞ p : n 1 maximize the reciprocal sine function 1/sin z. Hence, it is sufficient to { n ≥ } consider the infinite series over the numerators sequence p : n 1 . To accomplish this, { n ≥ } write the series as a sum of a convergent infinite series and the lacunary infinite series (over the sequence of numerators): 1 1 1 u v = u v + u v (95) n sin n m sin m q sin (απqn) n≥1 m≥1 n≥1 n X mX6=pn X 1 c0 u v . ≤ qn sin (απqn) nX≥1 where c0, c1, c2, c3, c4 > 0 are constants. By hypothesis, the reciprocal of the sine function is bounded by 1 c qa−1, (96) sin(απq ) ≤ 1 n n see Theorem 4.1 for more details. Applying this bound yields

(a−1)v 1 (c1qn) u v u qn sin (απqn) ≤ qn nX≥1 nX≥1 1 c . (97) 2 u−(a−1)v ≤ qn nX≥1 By the Binet formula for quadratic recurrent sequences, the sequence qn = anqn−1 + qn−2, a 1, has exponential growth, namely, n ≥ n 1 1+ √5 qn (98) ≥ √5 2 ! for n 2. Now, replacing (98) into (79) returns ≥ 1 2 (u−(a−1)v)n c c 2 u−(a−1)v 3 qn ≤ 1+ √5 Xn≥1 nX≥1 1 (u−(a−1)v)n c . (99) ≤ 4 2 nX≥1 Irrationality Measure of Pi 20

Hence, it immediately follows that the inﬁnite series converges whenever u (a 1)v > − − 0.

Example 12.1. The irrational number α = √2 has irrationality measure µ(√2) = 2, and the inﬁnite series 1 (100) n3 sin2 √2πn nX≥1 has u (a 1)v = 3 2 = 1 > 0. Hence, by Theorem 12.1, it is absolutely convergent. − − − 3 Example 12.2. Let ε> 0 be an arbitrary small number. The irrational number α = √2 3 has irrationality measure µ(√2) = 2, and the inﬁnite series 1 3 (101) n1+ε sin √2πn nX≥1 has u (a 1)v =1+ ε 1= ε> 0. Hence, by Theorem 12.1, it is absolutely convergent. − − − 13 Convergence Of The Flat Hills Series

Let x be the fractional part function and let α = min n α : n Z be the least { } || || {| − | ∈ } distance to the nearest integer. Another classes of problems arise from the other properties of the number π and other irrational numbers. One of these problems is the convergence of the Flat Hills series. Definition 13.1. Let a> 1 and b = 0 be a pair of real parameters. A Flat Hills series is 6 defined by an infinite sum of the following forms. 1 1 (i) , (iii) , na sinb πn na sinb πn nX≥1 || || nX≥1 { } 1 1 (ii) , (iv) . na sinb π10n na sinb π10n nX≥1 || || nX≥1 { } The known properties of some algebraic irrational numbers α R can be used to determine ∈ the convergence or divergence of the Flat Hills series 1 , (102) na sinb αn nX≥1 || || see Exercise 17.26. In the case α = π, the analytic properties of sequences such as πn : n 1 , and π10n : n 1 are unknown. Accordingly, the convergence or {|| || ≥ } {|| || ≥ } divergence of the infinite series (102) are unknown. Likewise, for the case α = e, the analytic properties of sequences such as en : n 1 , and e10n : n 1 are unknown. {|| || ≥ } {|| || ≥ } Accordingly, the convergence or divergence of the infinite series (102) are unknown, see Lemma 16.6 for some details.

14 Representations

The Flint Hills series and similar class of inﬁnite series have representations in terms of other analytic functions. These reformulations could be useful in further analysis of these series. Irrationality Measure of Pi 21

14.1 Gamma Function Representation The gamma function reﬂection formula produces a new reformulation as

1 Γ(1 n/π)vΓ(n/π)v = − , (103) nu sinv n nu nX≥1 Xn≥1 see Lemma 15.1.

14.2 Zeta Function Representation The zeta function reﬂection relation

ζ(s) = 2sπ1−sΓ(1 s) sin (πs/2) ζ(1 s) (104) − − evaluated at s = 2n/π produces a new reformulation as

v 1 22n/ππ1−2n/πΓ(1 2n/π)ζ(1 2n/π) = − − , (105) nu sinv n nuζ(2n/π)v n≥1 n≥1 X X 14.3 Harmonic Kernel Representation The harmonic summation kernel is deﬁned by

sin((2x + 1)z) (z)= ei2nz = , (106) Kx sin (z) −xX≤n≤x where x, z C are complex numbers. For z = kπ, the harmonic summation kernel has ∈ 6 the upper bound (z) x . Replacing it into the series produces a new reformulation |Kx | ≪ | | as 1 (n)v = Kx . (107) nu sinv n nu sinv((2x + 1)n) Xn≥1 nX≥1 Here the choice of parameter x = αn, where α> 0 is an algebraic irrational number, shifts the large value of the reciprocal sine function 1/sin n to (z), and the 1/sin(2x + 1)n = Kx sin(2αn + 1)n remains bounded. Thus, it easier to prove the convergence of the series (107).

15 Beta and Gamma Functions

1 a−1 b−1 Certain properties of the beta function B(a, b)= 0 t (1 t) dt, and gamma function ∞ − Γ(z)= tz−1e−tdt are useful in the proof of the main result. 0 R LemmaR 15.1. Let B(a, b) and Γ(z) be the beta function and gamma function of the complex numbers a, b, z C 0, 1, 2, 3,... . Then, ∈ − { − − − } (i) Γ(z +1) = zΓ(z), the functional equation. Γ(a)Γ(b) (ii) B(a, b)= , the multiplication formula. Γ(a + b) B(1 z, z) 1 (iii) − = . π sin πz Irrationality Measure of Pi 22

Γ(1 z)Γ(z) 1 (iv) − = , the reﬂection formula. π sin πz Proof. Standard analytic methods on the B(a, b), and gamma function Γ(z), see [11, Equation 5.12.1], [2, Chapter 1], et cetera.

The current result on the irrationality measure µ(π) = 7.6063, see Table 1, of the number π implies that

Γ(z + 1)Γ(z) z 7.6063. (108) | | ≪ | | A sharper upper bound is computed here.

Lemma 15.2. Let Γ(z) be the gamma function of a complex number z C, but not a ∈ negative integer z = 0, 1, 2,.... Then, 6 − − Γ(z + 1)Γ(z) z . (109) | | ≪ | | Proof. Assume e(z) > 0. Then, the functional equation ℜ Γ(z +1) = zΓ(z), (110) see Lemma 15.1, or [11, Equation 5.5.1], provides an analytic continuation expression

Γ(1 z)Γ(z) 1 − = , π sin πz for any complex number z C such that z = 0, 1, 2, 3,.... By Theorem 5.1, ∈ 6 − − − 1 1 sin πz ≍ 1 | | sin |πz| πz ≍ | | as πz . | |→ ∞ 16 Basic Diophantine Approximations Results

All the materials covered in this section are standard results in the literature, see [14], [17], [20], [24], [25], [28], et alii.

16.1 Rationals And Irrationals Numbers Criteria A real number α R is called rational if α = a/b, where a, b Z are integers. Otherwise, ∈ ∈ the number is irrational. The irrational numbers are further classiﬁed as algebraic if α is the root of an irreducible polynomial f(x) Z[x] of degree deg(f) > 1, otherwise it is ∈ transcendental.

Lemma 16.1. If a real number α R is a rational number, then there exists a constant ∈ c = c(α) such that c p α (111) q ≤ − q

holds for any rational fraction p/q = α. Speciﬁcally, c 1/b if α = a/b. 6 ≥ Irrationality Measure of Pi 23

This is a statement about the lack of effective or good approximations for any arbitrary rational number α Q by other rational numbers. On the other hand, irrational numbers ∈ α R Q have effective approximations by rational numbers. If the complementary ∈ − inequality α p/q < c/q holds for infinitely many rational approximations p/q, then it | − | already shows that the real number α R is irrational, so it is sufficient to prove the ∈ irrationality of real numbers.

Lemma 16.2 (Dirichlet). Suppose α R is an irrational number. Then there exists an ∈ inﬁnite sequence of rational numbers pn/qn satisfying

p 1 0 < α n < (112) − q q2 n n for all integers n N. ∈ Lemma 16.3. Let α = [a0, a1, a2,...] be the continued fraction of a real number, and let p /q : n 1 be the sequence of convergents. Then { n n ≥ } p 1 0 < α n < (113) − q a q2 n n+1 n for all integers n N. ∈ This is standard in the literature, the proof appears in [14, Theorem 171], [25, Corollary 3.7], [15, Theorem 9], and similar references.

Lemma 16.4. Let α = [a0, a1, a2,...] be the continued fraction of a real number, and let p /q : n 1 be the sequence of convergents. Then { n n ≥ } 1 p 1 1 p 1 (i) α n , (ii) α n , 2q q ≤ − q ≤ q2 2a q2 ≤ − q ≤ q2 n+1 n n n n+1 n n n for all integers n N. ∈ The recursive relation qn+1 = an+1qn+qn−1 links the two inequalities. Confer [21, Theorem 3.8], [15, Theorems 9 and 13], et alia. The proof of the best rational approximation stated below, appears in [24, Theorem 2.1], and [25, Theorem 3.8].

Lemma 16.5. Let α R be an irrational real number, and let p /q : n 1 be the ∈ { n n ≥ } sequence of convergents. Then, for any rational number p/q Q×, ∈ p p (i) αqn pn αq p , (ii) α n α , | − | ≤ | − | − q ≤ − q n for all suﬃciently large n N such that q q . ∈ ≤ n 16.2 Irrationalities Measures The concept of measures of irrationality of real numbers is discussed in [28, p. 556], [5, Chapter 11], et alii. This concept can be approached from several points of views. Irrationality Measure of Pi 24

Deﬁnition 16.1. The irrationality measure µ(α) of a real number α R is the inﬁmum ∈ of the subset of real numbers µ(α) 1 for which the Diophantine inequality ≥ p 1 α (114) − q ≪ qµ(α)

has ﬁnitely many rational solutions p and q. Equivalently, for any arbitrary small number ε> 0 p 1 α (115) − q ≫ qµ(α)+ε

for all large q 1. ≥ Theorem 16.1. ([7, Theorem 2]) The map µ : R [2, ) 1 is surjective function. −→ ∞ ∪ { } Any number in the set [2, ) 1 is the irrationality measure of some irrational number. ∞ ∪ { } Example 16.1. Some irrational numbers of various irrationality measures.

(1) A rational number has an irrationality measure of µ(α) = 1, see [14, Theorem 186].

(2) An algebraic irrational number has an irrationality measure of µ(α) = 2, an introduction to the earlier proofs of Roth Theorem appears in [24, p. 147].

(3) Any irrational number has an irrationality measure of µ(α) 2. ≥ (4) A Champernowne number κ = 0.123 b 1 b b + 1 b + 2 in base b 2, b · · · − · · · · · · ≥ concatenation of the b-base integers,

has an irrationality measure of µ(κb)= b.

n (5) A Mahler number ψ = b−[τ] in base b 3 has an irrationality measure of b n≥1 ≥ µ(ψ )= τ, for any real b P number τ 2, see [7, Theorem 2]. ≥ (6) A Liouville number ℓ = b−n! parameterized by b 2 has an irrationality b n≥1 ≥ measure of µ(ℓ )= , see [14, p. 208]. b ∞ P Deﬁnition 16.2. A measure of irrationality µ(α) 2 of an irrational real number α R× ≥ ∈ is a map ψ : N R such that for any p,q N with q q , → ∈ ≥ 0 p 1 α . (116) − q ≥ ψ(q)

Furthermore, any measure of irrationality of an irrational real number satisﬁes ψ(q) ≥ √5qµ(α) √5q2. ≥ Theorem 16.2. For all integers p,q N, and q q0, the number π satisﬁes the rational approximation inequality ∈ ≥ p 1 π . (117) − q ≥ q7.6063

Proof. Consult the original source [26, Theorem 1]. Irrationality Measure of Pi 25

16.3 Normal Numbers The earliest study was centered on the distribution of the digits in the decimal expansion of the number √2 = 1.414213562373 . . ., which is known as the Borel conjecture.

Deﬁnition 16.3. An irrational number α R is a normal number in base 10 if any ∈ sequence of k-digits in the decimal expansion occurs with probability 1/10k.

Lemma 16.6. (Wall) An irrational number α R is a normal number in base 10 if and ∈ only if the sequence α10n : n 1 is uniformly distributed modulo 1. { ≥ } 17 Problems

17.1 Flint Hill Class Series Exercise 17.1. The inﬁnite series ( 1)n+1 1 S = − and S = 1 n 2 n2 sin n Xn≥1 nX≥1 have many asymptotic similarities. The ﬁrst is conditionally convergent. Is the series S2 conditionally convergent?

Exercise 17.2. Show that the inﬁnite series cosv n sinv n S = and S = 3 n 4 n Xn≥1 nX≥1 are conditionally convergent for odd v 1. Are these series absolutely conditionally ≥ convergent for any v > 0?

Exercise 17.3. Explain the convergence of the inﬁnite series

π 1 ( 1)n = + 2z − sin πz z z2 n2 Xn≥1 − at the real numbers z = n/π, where n 1 is an integer. Is this series convergent? ≥ Exercise 17.4. Explain the convergence of the inﬁnite series 1 1 π cot πz = + 2z z z2 n2 Xn≥1 − at the real numbers z = n/π, where n 1 is an integer. Is this series convergent? The ≥ derivation of this series appears in [9, p. 122].

Exercise 17.5. Explain the convergence of the inﬁnite series π 1 1 = 2 π2 (z n)2 sin πz −∞

Exercise 17.6. Explain the convergence of the inﬁnite series sin z 2 ( 1)n sin nz = − sin πz π z2 n2 nX≥1 − at the real numbers z = n/π, where n 1 is an integer. This series has a diﬀerent ≥ structure than the previous exercise, is this series is absolutely convergent? conditionally convergent? The derivation of this series appears in [12, p. 276].

17.2 Complex Analysis Exercise 17.7. Assume that the infinite series is absolutely convergent. Use a Cauchy integral formula to evaluate the integral 1 1 dz, i2π zu sinv z ZC where C is a suitable curve. Exercise 17.8. Determine whether or not there is a complex valued function f(z) for which the Cauchy integral evaluate to 1 1 f(z)dz = u v , i2π C n sin n Z nX≥1 where C is a suitable curve, and the infinite series is absolutely convergent. Exercise 17.9. Let z C be a large complex variable. Prove or disprove the asymptotic ∈ relation 1 1 πz . sin πz ≍ 1 ≍ | | | | sin |πz| Exercise 17.10. Consider the product 1 sin sin z =1+ O(1/z2) (118) z × of a complex variable z C. Explain its properties in the unit disk z C : z < 1 and ∈ { ∈ | | } at infinity. Exercise 17.11. Let z C be a large complex variable. Use the properties of the gamma ∈ function to prove the asymptotic relation

Γ(1 z)Γ(z) = O( z ). | − | | | Exercise 17.12. Let x R be a large real variable. Use the trigamma reﬂection formula ∈ π2 ψ1(1 x)+ ψ1(x)= − sin2 (πx) to derive an asymptotic relation 1 πx 2. sin2 πx ≍ | | | | Explain any restrictions on the real variable x R. ∈ Irrationality Measure of Pi 27

17.3 Irrationality Measures Exercise 17.13. Do the irrationality measures of the numbers π and π2 satisfy µ(π) = µ(π2) = 2? This is supported by the similarity of sin(n) = sin(n mπ) and − sin x x2 = 1 x − π2m2 mY≥1 at x = n. More generally, x = nkπ−k+1

Exercise 17.14. What is the relationship between the irrationality measures of the numbers π and πk, for example, do these measures satisfy µ(π)= µ(πk)=2 for k 2? ≥ 17.4 Partial Quotients Exercise 17.15. Determine an explicit bound a B for the continued fraction of the n ≤ irrational number π = [a ; a , a , a ,...] for all partial quotients a as n . 0 1 2 3 n →∞ Exercise 17.16. Does π2 = [a ; a , a , a ,...] has bounded partial quotients a as n 0 1 2 3 n → ? ∞ Exercise 17.17. Let α = [a0; a1, a2, a3,...] and β = [b0; b1, b2, b3,...] be a pair of continued fractions. Is there an algorithm to determine whether or not α = β or α = β based 6 on sequence of comparisons a = b , a = b , a = b , . . . , a = b for some N 1? 0 0 1 1 2 2 N N ≥ Exercise 17.18. Let e = [a0; a1, a2, a3,...], where a3k = a3k+2 = 1, and a3k+1 = 2k, be the of continued fraction of the natural base e. Compute the geometric mean value of the partial quotients: 1/n

Ke = lim ak = ? n→∞   ∞ kY≤n   Exercise 17.19. Let π = [a0; a1, a2, a3,...] be the of continued fraction of the natural base. A short caculation give

1/n 1/n

Kπ = lim ak = 3.361 ..., Kπ = lim ak = 2.628 .... n≤10   n≤20   kY≤n kY≤n     Compute a numerical approximation for the geometric mean value of the partial quotients:

1/n

Kπ = lim ak =? n≤1000   kY≤n   Exercise 17.20. Let α = [a0; a1, a2, a3,...] be the of continued fraction of a real number.

Assume it has a lacunary subsequence of unbounded partial quotients ani = O(log log n), such that ni+1/ni > 1, otherwise an = O(1). Is the geometric mean value of the partial quotients unbounded 1/n

Kα = lim ak = ? n→∞   ∞ kY≤n   Irrationality Measure of Pi 28

17.5 Concatenated Sequences Exercise 17.21. A Champernowne number κ = 0.123 b 1 b b + 1 b + 2 in b · · · − · · · · · · base b 2 is formed by concatenating the sequence of consecutive integers in base b is ≥ irrationality. Show that the number 0.F0F1F2F3 . . . = 1/F11 formed by concatenating the sequence of Fibonacci numbers Fn+1 = Fn + Fn−1 is rational. Exercise 17.22. Let f(x) Z[x] be a polynomial, and let D = f(n) . Show that the ∈ n | | number 0.D0D1D2D3 . . . formed by concatenating the sequence of values is irrational. Exercise 17.23. Let D 0 : n 0 be an inﬁnite sequence of integers, and let { n ≥ ≥ } α = 0.D0D1D2D3 . . . formed by concatenating the sequence of integers. Determine a suﬃcient condition on the sequence of integers to have an irrational number α> 0.

17.6 Exact Evaluations Of Power Series Exercise 17.24. The exact evaluation of the ﬁrst series below is quite simple, but the next series requires some work: 1 1 π2 1 L (1/2) = = log 2 and L (1/2) = = (log 2)2 . 1 2nn 2 2nn2 12 − 2 Xn≥1 nX≥1 n k Prove this, use the properties of the polylogarithm function Lk(z)= n≥1 z /n . Exercise 17.25. Determine whether or not the series P 1 L (1/2) = 3 2nn3 nX≥1 has a closed form evaluation (exact).

17.7 Flat Hill Class Series Exercise 17.26. Let α = (1+ √5)/2 be a Pisot number. Show that αn 0 as n . || ||→ →∞ Explain the convergence of the first infinite series 1 sin αn and || || n2 sin αn Xn≥1 nX≥1 || || the divergence of the second infinite series. Exercise 17.27. Assume that π is a normal number base 10. Show that π10n is || || uniformly distributed. Explain the convergence or divergence of the infinite series 1 n2 sin π10n nX≥1 || || . Exercise 17.28. Assume that π is a normal number base 10. Determine the least parameters a> 1 and b> 0 for which the infinite series 1 na sinb πn nX≥1 || || converges. For example, is a b> 0 sufficient? − Irrationality Measure of Pi 29

Exercise 17.29. Assume that π is a normal number base 10. Determine the least parameters a> 1 and b> 0 for which the inﬁnite series 1 na sinb π10n Xn≥1 || || converges. For example, is a b> 0 suﬃcient? − References

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18 Numerical Data

The maxima of the function 1/sin x occur at the numerators x = pn of the sequence of convergents p /q π. The ﬁrst few terms of the sequence p , which is cataloged as n n −→ n A046947 in [22], are:

π = 1, 3, 22, 333, 355, 103993, 104348, 208341, 312689, 833719, 1146408, 4272943, N { (119) 5419351, 80143857, 165707065, 245850922, 411557987,... } Irrationality Measure of Pi 31

µ(π) Table 3: Numerical Data For Irrationality Measure p /q π qn . | n n − |≥ n pn qn µ(π) 1 3 1 2 22 7 3.429288 3 333 106 2.014399 4 355 113 3.201958 5 103993 33102 2.043905 6 104348 33215 2.096582 7 208341 66317 2.055815 8 312689 99532 2.107950 9 833719 265381 2.039080 10 1146408 364913 2.120203 11 4272943 1360120 2.020606 12 5419351 1725033 2.189381 13 80143857 25510582 2.057220 14 165707065 52746197 2.057220 15 245850922 78256779 2.040522 16 411557987 131002976 2.058941 17 1068966896 340262731 2.052652 18 2549491779 811528438 2.049381 19 6167950454 1963319607 2.057213 20 14885392687 473816765 2.203610 21 21053343141 6701487259 2.196409 22 1783366216531 567663097408 2.034261 23 3587785776203 1142027682075 2.032066 24 5371151992734 1709690779483 2.019525 25 8958937768937 2851718461558 2.096513

18.1 Data For The Irrationality Measure A few values were computed to illustrate the prediction in Theorem 1.1. The numerators pn and the denominators qn are listed in OEIS A002485 and A002486 respectively. The numerical data and Theorem 1.1 are very well matched. Irrationality Measure of Pi 32

Table 4: Numerical Data For 1/sin pn n pn 1/sin(pn) 1/sin 1/pn sin pn/sin(1/pn) 1 3 7.086178 3.0562 0.431303 2 22 112.978 22.0076 0.194796 − − 3 333 113.364 333.001 2.93745 − − 4 355 33173.7 355.0 0.0107013 − − 5 103993 52275.7 103993.0 1.98932 − − 6 104348 90785.1 104348.0 1.1494 − − 7 208341 123239.0 208341. 1.69055 8 312689 344744.0 312689.0 0.907017 9 833719 432354.0 833719.0 1.92832 10 1146408 1.70132 106 1.14641 106 0.673836 − × × − 11 4272943 1.81957 106 4.27294 106 2.34832 − × × − 12 5419351 2.61777 107 5.41935 106 0.207022 − × × − 13 80143857 6.76918 107 8.01439 107 1.18395 − × × − 14 165707065 1.15543 108 1.65707 108 1.43416 − × × − 15 245850922 1.6345 108 2.45851 108 1.50413 × × 16 411557987 3.9421 108 4.11558 108 1.04401 × × 17 1068966896 9.57274 108 1.06897 109 1.11668 × × 18 2549491779 2.23489 109 2.54949 109 1.14077 × × 19 6167950454 6.6785 109 6.16795 109 0.923554 × × 20 14885392687 6.75763 109 1.48854 1010 2.20275 × × 21 21053343141 5.70327 1011 2.10533 1010 0.0369145 × × 22 1783366216531 1.43483 1012 1.78337 1012 1.24291 × × 23 3587785776203 2.78176 1012 3.58779 1012 1.28975 × × 24 5371151992734 2.96328 1012 5.37115 1012 1.81257 − × × − 25 8958937768937 4.54136 1013 8.95894 1012 0.197274 − × × −

18.2 The Sine Asymptotic Expansions A few values were computed in Table 4 to illustrate the prediction in Theorem 5.1. Note that (86) is the restriction to convergents, so never vanishes, it is bounded below by 1/pn. The numerical data in Table 4 conﬁrms this result. Irrationality Measure of Pi 33

Table 5: Numerical Data For π/sin pn n p Γ(1 p /π)Γ(p /π) π2/p sin p n − n n n n 1 3 22.2619 23.3126 2 22 354.93 50.6838 − − 3 333 356.143 3.35992 − − 4 355 104218.0 922.286 − − 5 103993 164229.0 4.9613 − − 6 104348 285210.0 8.58678 − − 7 208341 387167.0 5.83812 8 312689 1.08305 106 10.8814 × 9 833719 1.35828 106 5.11823 × 10 1146408 5.34484 106 14.6469 − × − 11 4272943 5.71636 106 4.20283 − × − 12 5419351 8.22396 107 47.6742 − × − 13 80143857 2.1266 108 8.33615 − × − 14 165707065 3.62989 108 6.88181 − × − 15 245850922 5.13494 108 6.56166 × 16 411557987 1.23845 109 9.45359 × 17 1068966896 3.00736 109 8.83836 × 18 2549491779 7.02111 109 8.65171 × 19 6167950454 2.09811 1010 10.6866 × 20 14885392687 2.12297 1010 4.48058 × 21 21053343141 1.79173 1012 267.364 × 22 1783366216531 4.50764 1012 7.9407 × 23 3587785776203 8.73917 1012 7.65233 × 24 5371151992734 9.30941 1012 5.44508 − × − 25 8958937768937 1.42671 1014 50.0299 × −

18.3 The Sine Reﬂection Formula Observe that the substitution z p /π in Γ(1 z)Γ(z) leads to −→ n − Γ(1 z)Γ(z) π π2 − = = = O(1). (120) z z sin πz pn sin pn A few values were computed to illustrate the prediction in Lemma 15.2. The ratio (120) is tabulated in the third column of Table 5.